The adjoint operator of the operator L(u): d'u ;xE[0, 1] %3D dx2 with boundary conditions u(1)=2 u(0), and u'(1)=u'(0), is L(v) = - with boundary conditions d'v dx2 2 v(1) = v(0), and v (0)= v'(1). d'v L (v) = dx2 Ob. with boundary conditions v(1) = v(0), and v (0)=2 v'(1). d?v L(v) = dx2 with boundary conditions v(0)=2 v(1), and v'(0)= v'(1). O d. L(v) = - d'v with boundary conditions dx? v(1) = v(0), and v (0)=2 v'(1). d?v with boundary conditions Oe. = (^)+7 dx2 v(1) = v(0), and 2 v'(0)= v'(1). Of. d?v with boundary conditions dx2 =(^)+7 v(1)= 2 v(0), and v'(0)= v'(1).
The adjoint operator of the operator L(u): d'u ;xE[0, 1] %3D dx2 with boundary conditions u(1)=2 u(0), and u'(1)=u'(0), is L(v) = - with boundary conditions d'v dx2 2 v(1) = v(0), and v (0)= v'(1). d'v L (v) = dx2 Ob. with boundary conditions v(1) = v(0), and v (0)=2 v'(1). d?v L(v) = dx2 with boundary conditions v(0)=2 v(1), and v'(0)= v'(1). O d. L(v) = - d'v with boundary conditions dx? v(1) = v(0), and v (0)=2 v'(1). d?v with boundary conditions Oe. = (^)+7 dx2 v(1) = v(0), and 2 v'(0)= v'(1). Of. d?v with boundary conditions dx2 =(^)+7 v(1)= 2 v(0), and v'(0)= v'(1).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![d'u
dx2 XE10, 1]
The adjoint operator of the operator L(u) =
with boundary conditions u(1)=2 u(0), and u'(1)=u'(0),
is
d'v
with boundary conditions
O a.
L(v) = -
dx2
2 v(1) = v(0), and v'(0)= v'(1).
Ob.
= (^)T7
with boundary conditions
dx2
v(1) = v(0), and v (0)=2 v'(1).
d?v
with boundary conditions
L(v)=
dx2
v(0) = 2 v(1), and v'(0)= v'(1).
O d.
L(v)= -
d'v
with boundary conditions
dx?
v(1) = v(0), and v (0) = 2 v'(1).
O e.
=(^)+ 7
d'v
with boundary conditions
dx2
v(1) = v(0), and 2v (0)= v'(1).
d?v
L+(v)=
dx2
Of.
with boundary conditions
v(1) = 2 v(0), and v'(0)= v'(1).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b22b5bd-5b6c-46b5-91f9-e0569ad4ab63%2F1c0fc899-fd69-4c88-93dd-e69895e5533a%2Ferhuei_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d'u
dx2 XE10, 1]
The adjoint operator of the operator L(u) =
with boundary conditions u(1)=2 u(0), and u'(1)=u'(0),
is
d'v
with boundary conditions
O a.
L(v) = -
dx2
2 v(1) = v(0), and v'(0)= v'(1).
Ob.
= (^)T7
with boundary conditions
dx2
v(1) = v(0), and v (0)=2 v'(1).
d?v
with boundary conditions
L(v)=
dx2
v(0) = 2 v(1), and v'(0)= v'(1).
O d.
L(v)= -
d'v
with boundary conditions
dx?
v(1) = v(0), and v (0) = 2 v'(1).
O e.
=(^)+ 7
d'v
with boundary conditions
dx2
v(1) = v(0), and 2v (0)= v'(1).
d?v
L+(v)=
dx2
Of.
with boundary conditions
v(1) = 2 v(0), and v'(0)= v'(1).
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